# The Unapologetic Mathematician

## Eigenvectors of an Eigenpair

An eigenvalue $\lambda$ of a linear transformation $T$ is the same thing as a root of the characteristic polynomial of $T$. That is, the characteristic polynomial has a factor $(X-\lambda)$. We can evaluate this polynomial at $T$ to get the linear transformation $(T-\lambda I_V$. Vectors in the kernel of this space are the eigenvalues corresponding to the eigenvector $\lambda$.

Now we want to do the same thing with an eigenpair $(\tau,\delta)$. This corresponds to an irreducible quadratic factor $(X^2-\tau X+\delta)$ in the characteristic polynomial of $T$. Evaluating this polynomial at $T$ we get the transformation $T^2-\tau T+\delta I_V$, which I assert has a nontrivial kernel. Specifically, I want to focus in on some two-dimensional invariant subspace on which $T$ has no eigenvalues. This corresponds to a $2\times2$ block in an almost upper-triangular representation of $T$. So we’ll just assume for the moment that $V$ has dimension $2$.

What I assert is this: if $p$ is a monic polynomial (with leading coefficient ${1}$) of degree two, then either $p$ is the characteristic polynomial of $T$ or it’s not. If it is, then $p(T)=0$, and its kernel is the whole of $V$. If not, then the kernel is trivial, and $p(T)$ is invertible.

In the first case we can just pick a basis, find a matrix, and crank out the calculation. If the matrix of $T$ is $\displaystyle\begin{pmatrix}a&b\\c&d\end{pmatrix}$

then the characteristic polynomial is $X^2-(a+d)X+(ad-bc)$. We substitute the matrix into this polynomial to find \displaystyle\begin{aligned}&\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}-(a+d)\begin{pmatrix}a&b\\c&d\end{pmatrix}+(ad-bc)\begin{pmatrix}{1}&{0}\\{0}&{1}\end{pmatrix}=\\&\begin{pmatrix}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{pmatrix}-\begin{pmatrix}a^2+ad&ab+bd\\ac+cd&ad+d^2\end{pmatrix}+\begin{pmatrix}ad-bc&{0}\\{0}&ad-bc\end{pmatrix}=\\&\begin{pmatrix}{0}&{0}\\{0}&{0}\end{pmatrix}\end{aligned}

On the other hand, if $q$ is the characteristic polynomial and $p$ is any other monic polynomial of degree two, then $q(T)=0$, as we just showed. Then we can calculate $\displaystyle p(T)=p(T)-q(T)=aT+bI_V$

for some constants $a$ and $b$, at least one of which must be nonzero. If $a=0$, then $p(T)$ is a nonzero multiple of the identity, which is invertible as claimed. On the other hand, if $a\neq0$, then $\displaystyle p(T)=a(T-\frac{b}{a}I_V)$

which must be invertible since we assumed that $T$ has no eigenvalues.

So for any $2\times2$ block, the action of an irreducible quadratic polynomial in $T$ either kills off the whole block or has a trivial kernel. This makes it reasonable to define an eigenvector of the eigenpair $(\tau,\delta)$ to be a vector in the kernel $\mathrm{Ker}\left(T^2-\tau T+\delta I_V\right)$, in analogy with the definition of an eigenvector of a given eigenvalue.

April 3, 2009 - Posted by | Algebra, Linear Algebra